R is included in (R-{0} )x(R-{0} ) R = {(x,y) : xy >0} Show that R...

R is included in (R-{0} )x(R-{0} )

R = {(x,y) : xy >0}

Show that R is an equivalent relation and find f its equivalent classes

Expert Solution

Colby Messinger answered 2 years ago

(1-x)y" + xy' - y = 0 in power series

Determine whether each statement is true or false, and prove or disprove, as appropriate. (a) (∀x∈R)(∃y∈R)[xy=1].(∀x∈R)(∃y∈R)[xy=1]....

Determine whether each statement is true or false, and prove or disprove, as appropriate. (a) (∀x∈R)(∃y∈R)[xy=1].(∀x∈R)(∃y∈R)[xy=1]. (b) (∃x∈R)(∀y∈R)[xy=1].(∃x∈R)(∀y∈R)[xy=1]. (c) (∃x∈R)(∀y∈R)[xy>0].(∃x∈R)(∀y∈R)[xy>0]. (d) (∀x∈R)(∃y∈R)[xy>0].(∀x∈R)(∃y∈R)[xy>0]. (e) (∀x∈R)(∃y∈R)(∀z∈R)[xy=xz].(∀x∈R)(∃y∈R)(∀z∈R)[xy=xz]. (f) (∃y∈R)(∀x∈R)(∃z∈R)[xy=xz].(∃y∈R)(∀x∈R)(∃z∈R)[xy=xz]. (g) (∀x∈Q)(∃y∈Z)[xy∈Z].(∀x∈Q)(∃y∈Z)[xy∈Z]. (h) (∃x∈Z+)(∀y∈Z+)[y≤x].(∃x∈Z+)(∀y∈Z+)[y≤x]. (i) (∀y∈Z+)(∃x∈Z+)[y≤x].(∀y∈Z+)(∃x∈Z+)[y≤x]. (j) (∀x,y∈Z)[x<y⇒(∃z∈Z)[x<z<y]].(∀x,y∈Z)[x<y⇒(∃z∈Z)[x<z<y]]. (k) (∀x,y∈Q)[x<y⇒(∃z∈Q)[x<z<y]].(∀x,y∈Q)[x<y⇒(∃z∈Q)[x<z<y]].

Question

R is included in (R-{0} )x(R-{0} ) R = {(x,y) : xy >0} Show that R...

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